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Vlada [557]
3 years ago
13

Help needed I’m not good at geometry

Mathematics
1 answer:
TiliK225 [7]3 years ago
4 0

Answer:

I think it's 19.21, but in your case it would be 19.2. Sorry if I'm wrong

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Which of these could be a monomial?
masya89 [10]
Mono means one..but monomial can also a negative. Mono meaning one means there is a single variable. 
6 0
3 years ago
What is the answer to. Write a polynomial equation with integer coefficients that has the given roots, x=7 and x=-5
rewona [7]
\bf \begin{cases}
x=7\implies &x-7=0\\
x=-5\implies &x+5=0
\end{cases}\implies (x-7)(x+5)=\stackrel{original}{polynomial}
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x^2-2x-35=y\implies 1x^2-2x-35=y
7 0
3 years ago
Jess has 60 ounces of an alloy that is 40% gold. How many ounces of pure gold must be added to this alloy to create a new alloy
tia_tia [17]

Answer:

84 ounces of pure gold

Step-by-step explanation:

Jess has 60 ounces of an alloy that is 40% gold. How many ounces of pure gold must be added to this alloy to create a new alloy that is 75% gold?

Pure gold = 100% gold

Let the number of ounces of pure gold = x

Hence, we have the equation

40% × 60 ounces + 100%× x ounces = 75%(60 + x)ounces

= 0.4 × 60 + 1x = 0.75(60 + x)

= 24 + x = 45 + 0.75x

Collect like terms

x - 0.75x = 45 - 24

0.25x = 21

x = 21/0.25

x = 84 ounces

Therefore, we need 84 ounces of pure gold

6 0
3 years ago
PLEASE HELP HYPERBOLAS What is the length of the transverse axis ?<br> (x-1)^2/25-(y+3)^2/9=1
Sav [38]

Given equation of hyperbola is

\frac{(x-1)^{2}}{25}-\frac{(y+3)^{2}}{9}=1

Given hyperbola is the horizontal hyperbola

The standard form of the horizontal hyperbola is

\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1

When we compare these two equations, we get

h = 1, a = 5 , k = -3 and b = 3

Length of the transverse axis = 2a

Plug in the value of 'a' as 5

So,Length of the transverse axis = 2(5) = 10

7 0
3 years ago
How many different perfect cubes are among the positive actors of 2021^2021
9966 [12]

Answer:

hope this helps :D

Step-by-step explanation:

Perfect cube factors:

If a number is a perfect cube, then the power of the prime factors should be divisible by 3.

Example 1:Find the number of factors of293655118 that are perfect cube?

Solution: If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2(0 or 3 or 6or 9)—– 4 factors

3(0 or 3 or 6)  —–  3  factors

5(0 or 3)——- 2 factors

11(0 or 3 or 6 )— 3 factors

Hence, the total number of factors which are perfect cube 4x3x2x3=72

Perfect square and perfect cube

If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.

Example 2: How many factors of 293655118 are both perfect square and perfect cube?

Solution: If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have

2(0 or 6)—– 2 factors

3(0 or 6) —– 2 factors

5(0)——- 1 factor

11(0 or 6)— 2 factors

Hence total number of such factors are 2x2x1x2=8

Example 3: How many factors of293655118are either perfect squares or perfect cubes but not both?

Solution:

Let A denotes set of numbers, which are perfect squares.

If a number is a perfect square, then the power of the prime factors should be divisible by 2. Hence perfect square factors must have

2(0 or 2 or 4 or 6 or 8)—– 5 factors

3(0 or 2 or 4 or 6)  —– 4 factors

5(0 or 2or 4 )——- 3 factors

11(0 or 2or 4 or6 or 8 )— 5 factors

Hence, the total number of factors which are perfect square i.e. n(A)=5x4x3x5=300

Let B denotes set of numbers, which are perfect cubes

If a number is a perfect cube, then the power of the prime factors should be divisible by 3. Hence perfect cube factors must have

2(0 or 3 or 6or 9)—– 4 factors

3(0 or 3 or 6)  —–  3  factors

5(0 or 3)——- 2 factors

11(0 or 3 or 6 )— 3 factors

Hence, the total number of factors which are perfect cube i.e. n(B)=4x3x2x3=72

If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have

2(0 or 6)—– 2 factors

3(0 or 6) —– 2 factors

5(0)——- 1 factor

11(0 or 6)— 2 factors

Hence total number of such factors are i.e.n(A∩B)=2x2x1x2=8

We are asked to calculate which are either perfect square or perfect cubes i.e.

n(A U B )= n(A) + n(B) – n(A∩B)

=300+72 – 8

=364

Hence required number of factors is 364.

8 0
3 years ago
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